July 28th, 2017, 01:38 AM  #1 
Newbie Joined: Jul 2017 From: Iraq Posts: 18 Thanks: 0  Fourier Transform
Hello everyone, am trying to solve this Fourier Trans. problem, here is the original solution >> https://i.imgur.com/eJJ5FLF.png Q/ How did he come up with this result and where is my mistake? All equation are in the above attached picture here is my attempt, part 1>> https://i.imgur.com/DT2tJ0y.jpg part 2>> https://i.imgur.com/jopEoQd.jpg part 3>> https://i.imgur.com/cKXoekT.jpg 
July 28th, 2017, 02:48 AM  #2 
Senior Member Joined: Feb 2016 From: Australia Posts: 1,597 Thanks: 546 Math Focus: Yet to find out. 
You are out by a factor of $\omega$ in the final line, and perhaps consequently, a sign change. I suspect it was introduced in part 2 somewhere.

July 28th, 2017, 03:50 AM  #3  
Newbie Joined: Jul 2017 From: Iraq Posts: 18 Thanks: 0  second attempt Quote:
but also there is a little difference at the final results https://i.imgur.com/jkVdK7z.jpg and here is the original solution https://i.imgur.com/eJJ5FLF.png can you spot where is the mistake ??  
July 28th, 2017, 04:22 AM  #4 
Senior Member Joined: Feb 2016 From: Australia Posts: 1,597 Thanks: 546 Math Focus: Yet to find out. 
What did you do on 'line 5'

July 28th, 2017, 04:29 AM  #5 
Newbie Joined: Jul 2017 From: Iraq Posts: 18 Thanks: 0  No, I just write that word to remember it (not that important). In the previous solution there is (negative sign) which I forgot to include, but now I've included it, and, as you can see, the solution is somehow still different from the original one. Can you explain why? Last edited by skipjack; July 28th, 2017 at 05:50 AM. 
July 28th, 2017, 05:46 AM  #6 
Global Moderator Joined: Dec 2006 Posts: 19,051 Thanks: 1618 
$\displaystyle \int_0^\pi\! e^{j\omega t}\sin(t)dt = \left[\frac{e^{j\omega t}(\cos(t) + j\omega\sin(t))}{1  \omega^2}\right]_0^\pi = \frac{1 + e^{j\omega\pi}}{1  \omega^2}$. In the "original" solution, $\theta$ should be $t$. 
July 28th, 2017, 06:56 AM  #7  
Newbie Joined: Jul 2017 From: Iraq Posts: 18 Thanks: 0  question Quote:
Can you elaborate more? And is it correct that he reversed the integrals limit in the original solution? Last edited by skipjack; July 28th, 2017 at 06:59 AM.  
July 28th, 2017, 07:08 AM  #8 
Global Moderator Joined: Dec 2006 Posts: 19,051 Thanks: 1618 
For that type of integral, see this article. The limits were swapped to compensate for omitting $j^2$ from the denominators. 
July 28th, 2017, 07:11 AM  #9 
Newbie Joined: Jul 2017 From: Iraq Posts: 18 Thanks: 0 
Thanks indeed for your article link, but can you explain more about the results of the integration?
Last edited by skipjack; July 28th, 2017 at 08:13 AM. 
July 28th, 2017, 08:16 AM  #10 
Global Moderator Joined: Dec 2006 Posts: 19,051 Thanks: 1618 
What specific thing are you referring to?


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